# $T\lceil R \rceil _{eA} \lfloor S \rceil$on

The answer is a 9-letter word

$$log_2(26)$$ $$\frac{1}{6} (\sqrt{595}-19)$$ $$-\frac{12^2}{100}$$ $$0$$
$$-\sqrt{\pi}$$ $$6\pi\cdot\tanh^2(2)$$ $$\frac{307819}{62500}$$ $$0$$
$$-\frac{8}{3}-\frac{3}{\pi}+\frac{5\pi}{2}$$ $$-\frac{6551\pi}{6921}$$ $$5+12γ$$ $$0$$
$$2^2\cdot5$$ $$L_6$$ $$\frac{1337}{42}\cdot\frac{666}{21201}$$ $$1$$

The empty header of the table has no significance, it is required for it to be formatted as a table (AFAIK).

Hint 1:

Don't mull over the terms in the expressions, their result value is important.
$$\lfloor \rceil$$ means round(), midpoint behavior is irrelevant for this puzzle.

Hint 2:

This requires specific . It is googlable with the right keywords, but even then you need to have a rough idea what this is about.

Some things to analyze: C4, R4C1 to R4C3, size of the grid.

Hint 3:

A 3x3 result will not hold the answer by itself. If you're stuck, just let all parts of this thing decay for a while...

• Is $Log_2$ instead of $\log_2$ intentional? And does the empty title line have any significance?
– chtz
Commented Jul 25 at 15:20
• @chtz See latest edit Commented Jul 25 at 18:40
• What's L_6 supposed to be Commented Jul 31 at 7:51
• @uggupuggu, I thought it could be en.wikipedia.org/wiki/Lp_space ($L^6$ space) but it is in index instead of exponent...
– JKHA
Commented Aug 1 at 4:59

Partial answer: I think I know where this is going, but there are still some loose ends that need tying up. Perhaps it's not worth writing this as an answer, but it's too long for a comment, so here goes ...

The bottom row:

The three expressions all yield the whole numbers 20, 18 and 1. (I think that L₆ is the sixth Lucas number.) If we use the not very mathematical, but very puzzly trick of converting those numbers to letters via A1/Z26, we get TRA.

The right column:

Now the whole table with its formulas looks a lot like a matrix, and TRA suggests transposition. If we transpose the matrix, we get {0, 0, 0, 1} in the last row.

This looks like a general transformation matrix where the upper left 3×3 block describes rotation and scaling and the upper right 1×3 vector describes a translation (after rotation and scaling). Multiplying this matrix with a vector {x, y, z, 1} yields a transformed vector {x′, y′, z′, 1}.

The title confirms that: The three capital letters are T, R and S and in computer graphics, the general translation matrix is often called TRS matrix for Translation, Rotation and Scaling.

The upper left 3×3 cells:

The first hint tells us not to find meaning in the expressions proper. The values are important, so I've just blindly typed them in. The transposed matrix is:

  4.7004397  -1.7724539   4.2323853            20.0
 0.89877031   17.517819  -2.9736416            18.0
      -1.44    4.925104   11.926588             1.0

        0.0         0.0         0.0             1.0

(I've assumed that the γ in A₃₃ is Euler's constant.)

The title:

The overall transformation matrix C can be divided into separate parts: a scaling matrix S, a rotation matrix R and a translation matrix T. The order of transformations (scaling, rotation, tralslation) is important.

Sequences of transformations can be represented by multiplying transformation matrices. Matrix multiplication is not commutative and later transformations must be multplied "from the left": C = C₃·C₂·C₁.

The overall transformation matrix is therefore C = T·R·S. And that's the order in which those capital letters appear in the title.

(There's also a celing operation on ⌈R⌉ and a rounding operation on ⌊S⌉, which are important, as hint 1 indicates.)

Now what?

I think it's clear where this is going: We need a nine-letter word and we can get nine numbers from the coordinates, i.e. the top 3×3 block, of a transformed matrix.

Because we don't have any other input to work on, I assume that we must transform the identity matrix. (But the input could also be the original, untransposed matrix.)

We probably must apply the A1/Z26 thing again, so we need whole numbers. We can get them by applying the ceiling and rounding operators. Now there are two ways to go about this:

Round the intermediate results
Scale the initial matrix; round the transformed matrix; rotate; ceil the result of that; finally translate. That yields:

 25    20    33
 23    36    18
  1    19    14

The numbers 36 and 33 are out of bounds for the A1/Z26 approach and the rest don't make useful letters. I'm not sure that this is what the title means, because if so the round/ceil braces would have to be nested.

Round the matrices
Scale, then rotate, then translate. Instead of T·R·S use T·⌈R⌉·⌊S⌉, where the rounding and ceiling operations are applied to all elements of the respective transformation matrices. The result is:

 25    19    25
 19    36    16
  0     6    13

There's a zero and a 36, which are out of range for the A1/Z26 conversion. Applying the ceiling operator to the rotation matrix gives a matrix that is no longer a pure rotation, that is a matrix whose determinant isn't 1. Still, this interpretation seems to the the closest to what the title represents.

There's still a mistake somewhere. Is the starting matrix something else, perhaps the original matrix? Is there a better way to apply the roundings?

I've said earlier that I half-expect the anser to be TREACHERY, a synonym to treason. Now I think the answer might be TRANSFORM. Both answers start with TR, which are already there in the translation matrix.

• (I've found a better interpretation of the title, but I need to look into it further.) Commented Aug 1 at 12:57
• rot13(Lbh ner ba gur evtug genpx, bs pbhefr. Cyrnfr abgr gung V znqr na nfvavar zvfgnxr va gur svany nafjre, jvgu bar yrggre orvat -10 bs jung vg vf fhccbfrq gb or. Lbhe pbzzrag fbhaqf cebzvfvat, znlor sbphf ba gur rnfvre cneg svefg, vg'f rabhtu gb trg gur nafjre orlbaq ernfbanoyr qbhog juvyr xrrcvat zl zvfgnxr va zvaq.) Commented Aug 1 at 16:57
• Thanks for your feedback, Lukas. rot13(Gur vqrn V uvagrq ng va zl pbzzrag vf gung jr fubhyq frcnengr gur G, E naq F cnegf naq nccyl gurz va gur tvira beqre: G, E, prvy gur erfhyg, gura F, ebhaq gur erfhyg. (Be creuncf prvy gur ryrzragf bs E naq ebhaq gur barf bs F.) V unira'g orra fhprffshy, rira vs V trg n cerggl begubabezny E zngevk. V guvax V pna thrff jung gur nafjre vf -- GENAFSBEZ, naq gur svefg guerr ner nyernql gurer vs gur ynfg fgrc vf gb genafcbfr ntnva.) That's it for today. I'll go to bed now. Commented Aug 1 at 20:18
• A bit closer, I think, but still not there. For what it's worth, here's the code I've used. Commented Aug 2 at 7:04
• Thanks. I guess that decay hints at the decomposition of the matrix. I've tried that, but to no avail. I'm not getting anywhere. Perhaps someone else can solve this. Commented Aug 2 at 12:16